%%

The differential equation of the family of circles passing through the points $$(0,2)$$ and $$(0,-2)$$ is :

JEE · Math · previous-year question

  1. A.$$2 x y \frac{d y}{d x}+\left(x^{2}-y^{2}+4\right)=0$$correct
  2. B.$$2 x y \frac{d y}{d x}+\left(x^{2}+y^{2}-4\right)=0$$
  3. C.$$2 x y \frac{d y}{d x}+\left(y^{2}-x^{2}+4\right)=0$$
  4. D.$$2 x y \frac{d y}{d x}-\left(x^{2}-y^{2}+4\right)=0$$

Answer

A. $$2 x y \frac{d y}{d x}+\left(x^{2}-y^{2}+4\right)=0$$

Explanation

Family of circles passing through the points (0, 2) and (0, $$-$$2) $${x^2} + (y - 2)(y + 2) + \lambda x = 0,\,\lambda \in R$$ $${x^2} + {y^2} + \lambda x - 4 = 0$$ ...... (1) Differentiate w.r.t x $$2x + 2y{{dy} \over {dx}} + \lambda = 0$$ ....... (2) Using (1) and (2), eliminate $$\lambda$$ $${x^2} + {y^2} - \left( {2x + 2y{{dy} \over {dx}}} \right)x - 4 = 0$$ $$2xy{{dy} \over {dx}} + {x^2} - {y^2} + 4 = 0$$

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions